CodeForces - 608 B - Hamming Distance Sum
Genos needs your help. He was asked to solve the following programming problem by Saitama:
The length of some string s is denoted |s|. The Hamming distance between two strings s and t of equal length is defined as , where si is the i-th character of s and ti is the i-th character of t. For example, the Hamming distance between string "0011" and string "0110" is |0 - 0| + |0 - 1| + |1 - 1| + |1 - 0| = 0 + 1 + 0 + 1 = 2.
Given two binary strings a and b, find the sum of the Hamming distances between a and all contiguous substrings of b of length |a|.
Input
The first line of the input contains binary string a (1 ≤ |a| ≤ 200 000).
The second line of the input contains binary string b (|a| ≤ |b| ≤ 200 000).
Both strings are guaranteed to consist of characters '0' and '1' only.
Output
Print a single integer — the sum of Hamming distances between a and all contiguous substrings of b of length |a|.
Example
01
00111
3
0011
0110
2
Note
For the first sample case, there are four contiguous substrings of b of length |a|: "00", "01", "11", and "11". The distance between "01" and "00" is |0 - 0| + |1 - 0| = 1. The distance between "01" and "01" is |0 - 0| + |1 - 1| = 0. The distance between "01" and "11" is |0 - 1| + |1 - 1| = 1. Last distance counts twice, as there are two occurrences of string "11". The sum of these edit distances is 1 + 0 + 1 + 1 = 3.
The second sample case is described in the statement.
用第一串a对第二串b(b中长度为len(a)的子串)不断进行比较,数字相同答案不变,不同则答案加一。
若从一般的方式直接比较,显然会超时,但注意只有0和1两种状态,所以,只要记录a串的0的数量和1的数量,是可以避免对b串的重复操作。
#include<iostream> #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> #include<stack> #include<map> using namespace std; const int MAX=2e5+10; const double MIN=1e-6; #define INF 0x7fffffff #define ll long long #define FOR(i,n) for(i=1;i<=n;i++) #define mst(a) memset(a,0,sizeof(a)) #define mstn(a,n) memset(a,n,sizeof(a)) //struct num{int a,b;}a[1005]; //bool cmp(const num &x, const num &y){return x.a>y.a;} int main() { string a,b; int i; cin>>a; cin>>b; ll ans=0; int d=b.size()-a.size()+1; ll s=0; for(i=0;i<d;i++) { s+=b[i]-'0'; } for(i=0;i<a.size();i++) { if(a[i]=='0') ans+=s; else ans+=d-s; s+=(b[d+i]-'0')-(b[i]-'0'); } cout<<ans<<endl; return 0; }